IB CIE Physics: Intensive Calculation Practice | IB CIE 物理：计算题专项训练

📚 IB CIE Physics: Intensive Calculation Practice | IB CIE 物理：计算题专项训练

Mastering calculation questions in IB and CIE Physics requires a systematic approach, fluency in formula manipulation, and a deep understanding of how concepts link together. This article presents intensive practice drills targeting the most common problem types across the syllabus — from kinematics to nuclear physics — with paired explanations in English and Chinese to strengthen your bilingual problem-solving skills. Each section includes key equations, worked examples, and step-by-step reasoning to build the confidence you need for both school assessments and final examinations.

在IB和CIE物理中攻克计算题，需要系统的训练、纯熟的公式运用能力以及对概念之间联系的深刻理解。本文针对大纲中最常见的计算题型——从运动学到核物理——提供强化练习，并配以中英双语对照讲解，帮助你提升双语解题能力。每一节都包含核心公式、典型例题和分步推理，为校内评估和最终大考做好充分准备。

1. Kinematics Calculations | 运动学计算

Kinematics problems often involve the four SUVAT equations that describe uniformly accelerated motion along a straight line. Choosing the right equation depends on identifying which variables are known and which one you need to find. Always begin by listing the given quantities: initial velocity u, final velocity v, acceleration a, displacement s and time t.

运动学问题通常涉及描述匀加速直线运动的四个SUVAT方程。选择正确的方程取决于确定已知量和待求量。解题时总是先列出已知物理量：初速度u、末速度v、加速度a、位移s和时间t。

v = u + at   s = ut + ½at²   v² = u² + 2as   s = ½(u + v)t

Whenever an object is dropped or thrown near the Earth’s surface, the acceleration due to gravity g = 9.81 m s⁻² acts downwards. You must define a sign convention — usually downwards is positive or upwards is positive — and apply it consistently to all vectors.

当物体在地球表面附近下落或被抛掷时，重力加速度g = 9.81 m s⁻²朝下作用。必须规定一个正方向（通常以向下为正或以向上为正），并将此符号规则一致地应用于所有矢量。

Worked Example: A stone is thrown vertically upwards with a speed of 15 m s⁻¹ from the top of a cliff 20 m above the sea. Calculate the time taken for the stone to hit the water, taking upwards as positive.

例题：一块石头从海面上方20米的悬崖顶端以15 m s⁻¹的速度竖直向上抛出。取向上为正方向，计算石头落到水面所需的时间。

Step 1 English: List the known quantities with proper signs: u = +15 m s⁻¹, a = –9.81 m s⁻², s = –20 m (the water is 20 m below the starting point, so displacement is negative). We need to find t.

步骤1中文：列出带符号的已知量：u = +15 m s⁻¹，a = –9.81 m s⁻²，s = –20 m（水面在出发点下方20 m，因此位移为负）。我们需要求t。

Step 2 English: Choose the SUVAT equation that includes s, u, a and t: s = ut + ½at². Substitute the values: –20 = 15t + ½(–9.81)t² → –20 = 15t – 4.905t². Rearrange into a quadratic equation: 4.905t² – 15t – 20 = 0.

步骤2中文：选择包含s、u、a和t的SUVAT方程：s = ut + ½at²。代入数值：–20 = 15t + ½(–9.81)t² → –20 = 15t – 4.905t²。整理成二次方程：4.905t² – 15t – 20 = 0。

Step 3 English: Solve the quadratic using the formula t = [–b ± √(b² – 4ac)] / 2a with a = 4.905, b = –15, c = –20. The positive root gives t ≈ 3.67 s. The stone hits the water after about 3.67 seconds.

步骤3中文：用二次公式 t = [–b ± √(b² – 4ac)] / 2a 求解，其中 a = 4.905，b = –15，c = –20。取正根得到 t ≈ 3.67 s。因此石头大约3.67秒后落入水中。

---

2. Newton’s Laws and Friction | 牛顿定律与摩擦力

Newton’s second law states that the resultant force on an object equals its mass times its acceleration: F_net = ma. The first step in any dynamics problem is to draw a free‑body diagram showing all forces, then resolve them along convenient axes — usually parallel and perpendicular to the surface of contact.

牛顿第二定律指出，物体所受的合外力等于其质量乘以加速度：F_net = ma。任何动力学问题的第一步都是画出受力示意图，然后将各力沿合适的坐标系分解——通常沿接触面的平行和垂直方向。

Frictional forces behave in two distinct ways: static friction prevents motion up to a maximum value f_s,max = μ_s N, while kinetic friction acts during motion and equals f_k = μ_k N. Always check whether the system is moving to decide which coefficient to use.

摩擦力有两种不同表现：静摩擦力阻止物体相对滑动，最大值为 f_s,max = μ_s N；而动摩擦力在物体滑动时起作用，大小为 f_k = μ_k N。解题时必须先判断系统是否运动，以决定使用哪个摩擦系数。

Worked Example: A 5.0 kg block rests on a rough slope inclined at 30° to the horizontal. The coefficient of static friction is 0.60. Determine whether the block slides. If it does, calculate its acceleration given μ_k = 0.40.

例题：一个5.0 kg的物块放在倾角为30°的粗糙斜面上，静摩擦系数为0.60。判断物块是否会滑动。如果滑下，已知动摩擦系数为0.40，求其加速度。

Step 1 English: Resolve weight into components: mg sin 30° down the slope = 5.0 × 9.81 × 0.5 = 24.53 N, perpendicular component mg cos 30° = 5.0 × 9.81 × 0.866 = 42.48 N. The normal force N equals the perpendicular component.

步骤1中文：将重力分解：沿斜面向下的分量 mg sin 30° = 5.0 × 9.81 × 0.5 = 24.53 N，垂直斜面的分量 mg cos 30° = 5.0 × 9.81 × 0.866 = 42.48 N。法向力N等于垂直分量。

Step 2 English: Maximum static friction f_s,max = μ_s N = 0.60 × 42.48 = 25.49 N. Since the downhill component (24.53 N) is less than 25.49 N, the block does not slide — it remains stationary.

步骤2中文：最大静摩擦力 f_s,max = μ_s N = 0.60 × 42.48 = 25.49 N。由于下滑分力（24.53 N）小于25.49 N，物块不会滑动，保持静止。

---

3. Work, Energy and Power | 功、能量与功率

Work done by a constant force is W = F d cos θ, where θ is the angle between the force and displacement. The work–energy theorem states that net work equals the change in kinetic energy: W_net = ΔKE = ½mv² – ½mu². These principles allow you to bypass time when solving many dynamics problems.

恒力做的功为 W = F d cos θ，其中θ为力与位移的夹角。功–能定理指出，合外力的功等于动能的变化：W_net = ΔKE = ½mv² – ½mu²。利用这些原理可以在许多动力学问题中避开时间变量。

Power is the rate of doing work or transferring energy: P = W / t = F v cos θ. For vehicles moving against resistive forces, the useful power output at a given speed helps determine maximum acceleration or gradient capability.

功率是做功或能量转换的速率：P = W / t = F v cos θ。对于克服阻力行驶的车辆，在给定速度下的有用输出功率可用于确定最大加速度或爬坡能力。

Worked Example: A 1200 kg car accelerates from 10 m s⁻¹ to 25 m s⁻¹ on a level road against a constant resistive force of 600 N. If the engine provides a constant 45 kW of useful power, calculate the minimum time for this speed change.

例题：一辆1200 kg的汽车在水平路面上从10 m s⁻¹加速到25 m s⁻¹，受到恒定的600 N阻力。若发动机提供恒定的45 kW有用功率，计算实现这一速度变化的最短时间。

Step 1 English: First find the change in kinetic energy: ΔKE = ½ × 1200 × (25² – 10²) = 600 × (625 – 100) = 315 000 J.

步骤1中文：先求动能的变化量：ΔKE = ½ × 1200 × (25² – 10²) = 600 × (625 – 100) = 315 000 J。

Step 2 English: The work done by the engine must provide the KE increase plus overcome the resistive force over the distance moved. Using power P = (F_engine) v_avg, we can also apply the work–energy theorem considering net work: total work done by engine W_engine = ΔKE + work against resistance. But a more direct route: power × time = total energy required minus the loss to resistance? Instead, we can compute net driving force: F_drive = P / v (instantaneous), but this is variable. Alternatively, we can use the fact that time = (increase in KE + work vs resistance) / power, but we do not know distance. Use another method: the resistive force is constant, so the net work = ΔKE + F_resist × d. However, we can use impulse-momentum? Better: use average velocity to find distance. Average v = (10+25)/2 = 17.5 m s⁻¹. Time t = d / 17.5. But we have unknown d. We can combine: work done by engine = P × t = ΔKE + 600 × d. And d = v_avg × t = 17.5 t. Thus 45000 t = 315000 + 600 × 17.5 t → 45000 t = 315000 + 10500 t → (45000 – 10500) t = 315000 → 34500 t = 315000 → t = 9.13 s.

步骤2中文：发动机做的功 = 功率 × 时间 = 动能增量 + 克服阻力做的功。位移 d = 平均速度 × 时间 = 17.5 t。因此：45000 t = 315000 + 600 × 17.5 t → 45000 t = 315000 + 10500 t → 34500 t = 315000 → t ≈ 9.13 s。所以最短时间约为9.13秒。

---

4. Momentum and Impulse | 动量与冲量

The momentum of an object is p = m v, and the impulse delivered by a net force equals the change in momentum: F_net Δt = Δp = m v – m u. This relationship is invaluable for collisions and explosions, where forces act over very short time intervals. In isolated systems, total momentum is conserved.

物体的动量为 p = m v，合外力的冲量等于动量的变化：F_net Δt = Δp = m v – m u。这一关系在作用时间极短的碰撞和爆炸问题中极具价值。在孤立系统中，总动量守恒。

For collisions, always identify the type — elastic or inelastic. In perfectly inelastic collisions, objects stick together and kinetic energy is not conserved, but momentum is. For two‑body interactions, use m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, taking care with velocity signs.

处理碰撞问题时，要先判断碰撞类型——弹性或非弹性。在完全非弹性碰撞中，物体粘在一起，动能不守恒，但动量守恒。对于两体相互作用，使用 m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂，注意速度的正负号。

Worked Example: A 0.50 kg trolley moving at 4.0 m s⁻¹ collides head‑on with a stationary 1.5 kg trolley. After the collision, the lighter trolley rebounds at 1.2 m s⁻¹ in the opposite direction. Calculate the velocity of the heavier trolley after the collision and the coefficient of restitution.

例题：一辆0.50 kg的小车以4.0 m s⁻¹ 的速度与一辆静止的1.5 kg小车发生正碰。碰撞后，轻车以1.2 m s⁻¹ 的速度反向弹回。求重车碰撞后的速度以及恢复系数。

Step 1 English: Adopt the initial direction of the light trolley as positive. Momentum before: p_initial = 0.50×4.0 + 1.5×0 = 2.0 kg m s⁻¹. After collision, v_light = –1.2 m s⁻¹ (rebound). Let v_heavy be the unknown. Conservation: 2.0 = 0.50×(–1.2) + 1.5 v_heavy → 2.0 = –0.6 + 1.5 v_heavy → v_heavy = 2.6/1.5 = 1.73 m s⁻¹ (positive direction).

步骤1中文：以轻车初始方向为正。碰前动量：p_initial = 0.50×4.0 + 1.5×0 = 2.0 kg m s⁻¹。碰后轻车速度 v_轻 = –1.2 m s⁻¹（反弹）。设重车速度为 v_重。动量守恒：2.0 = 0.50×(–1.2) + 1.5 v_重 → 2.0 = –0.6 + 1.5 v_重 → v_重 = 1.73 m s⁻¹（正方向）。

Step 2 English: Relative speed of approach = 4.0 m s⁻¹. Relative speed of separation = v_heavy – (–1.2) = 1.73 + 1.2 = 2.93 m s⁻¹. Coefficient of restitution e = separation speed / approach speed = 2.93 / 4.0 = 0.73.

步骤2中文：接近相对速度 = 4.0 m s⁻¹。分离相对速度 = v_重 – (–1.2) = 1.73 + 1.2 = 2.93 m s⁻¹。恢复系数 e = 分离速度 / 接近速度 = 2.93 / 4.0 = 0.73。

---

5. Circular Motion and Gravitation | 圆周运动与万有引力

An object moving in a circle of radius r at constant speed v experiences a centripetal acceleration a_c = v²/r directed towards the centre. The centripetal force required is F_c = m v²/r = m ω² r, where ω = v/r. This force is always supplied by real forces such as tension, friction, gravity or the normal reaction.

物体以恒定速率v在半径为r的圆上运动时，会受到向心加速度 a_c = v²/r 指向圆心。所需的向心力为 F_c = m v²/r = m ω² r，其中 ω = v/r。这一向心力总是由真实力提供，如张力、摩擦力、重力或法向反作用力。

Newton’s law of gravitation gives the attractive force between two point masses: F = G M m / r². In satellite motion, this gravitational force provides the centripetal force, leading to relationships such as v = √(GM/r) and T² ∝ r³ (Kepler’s third law).

牛顿万有引力定律给出两点质量之间的吸引力：F = G M m / r²。在卫星运动中，这一引力提供向心力，从而导出关系式，如 v = √(GM/r) 和 T² ∝ r³（开普勒第三定律）。

Worked Example: A 1500 kg car travels over a hump‑backed bridge of radius of curvature 40 m. At what speed will the car lose contact with the road?

例题：一辆1500 kg的汽车驶过曲率半径为40 m的拱形桥顶。汽车在多大速度时将与路面失去接触？

Step 1 English: At the top of the bridge, the forces on the car are weight mg downwards and normal reaction N upwards. The net force towards the centre is mg – N = m v²/r. Loss of contact occurs when N = 0.

步骤1中文：在桥顶，汽车受重力 mg 向下和法向反力 N 向上。指向圆心的净力为 mg – N = m v²/r。当 N = 0 时，汽车失去接触。

Step 2 English: Set N = 0: mg = m v²/r → v = √(g r) = √(9.81 × 40) = √392.4 ≈ 19.8 m s⁻¹ (about 71 km h⁻¹).

步骤2中文：令 N = 0：mg = m v²/r → v = √(g r) = √(9.81 × 40) = √392.4 ≈ 19.8 m s⁻¹（约71 km h⁻¹）。

---

6. Simple Harmonic Motion and Waves | 简谐运动与波

In simple harmonic motion (SHM), the restoring force is proportional to the displacement and opposite in direction: F = –k x. The period of a mass–spring system is T = 2π √(m/k), while for a simple pendulum T = 2π √(L/g). These formulas are independent of amplitude for small oscillations.

在简谐运动（SHM）中，回复力与位移成正比且方向相反：F = –k x。弹簧振子的周期为 T = 2π √(m/k)，单摆的周期为 T = 2π √(L/g)。对于小角度摆动，这些公式与振幅无关。

Wave calculations link speed, frequency and wavelength through v = f λ. When waves pass from one medium to another, frequency remains unchanged while speed and wavelength change. Standing waves on strings or in pipes involve nodes and antinodes, with boundary conditions fixing the harmonic wavelengths.

波的计算通过 v = f λ 将波速、频率和波长联系起来。波从一种介质进入另一种介质时，频率保持不变，波速和波长发生变化。弦或管中的驻波涉及波节和波腹，边界条件决定了谐振波长。

Worked Example: A 0.200 kg mass attached to a spring oscillates with an amplitude of 0.050 m and a maximum speed of 0.80 m s⁻¹. Calculate the spring constant k and

Published by TutorHao | IB Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）